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Brownian motion on a circle can be generated by $\left(\cos\left(B_t\right),\sin\left(B_t\right)\right)$ where $B$ is Brownian motion on the real line. My question is what SDE was solved to get this as the solution?

user337663
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1 Answers1

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$$g(t,x)={{e}^{ix}}=(\cos x,\sin x)$$ $${{y}_{t}}=({{y}_{1}},{{y}_{2}})$$ $${{y}_{t}}={{e}^{i{{B}_{t}}}}$$ $$g(t,x)={{e}^{ix}}\xrightarrow{x={{B}_{t}}}{{e}^{i{{B}_{t}}}}=(\cos {{B}_{t}},\sin {{B}_{t}})$$ $$\xrightarrow{{{y}_{t}}=({{y}_{1}},{{y}_{2}})}{{y}_{1}}=\cos {{B}_{t}},\overset{{}}{\mathop{{}}}\,{{y}_{2}}=\sin {{B}_{t}}$$

$$d{{y}_{t}}=\frac{\partial g}{\partial t}{{|}_{(t,{{B}_{t}})}}dt+\frac{\partial g}{\partial x}{{|}_{(t,{{B}_{t}})}}d{{B}_{t}}+\frac{1}{2}\frac{{{\partial }^{2}}g}{\partial {{x}^{2}}}{{|}_{(t,{{B}_{t}})}}{{(d{{B}_{t}})}^{2}}$$ $$0dt+(-\sin {{B}_{t}},\cos {{B}_{t}})d{{B}_{t}}+\frac{1}{2}(-\cos {{B}_{t}},-sin{{B}_{t}}){{(d{{B}_{t}})}^{2}}$$ $$=0dt+(-\sin {{B}_{t}},\cos {{B}_{t}})d{{B}_{t}}+\frac{1}{2}(-\cos {{B}_{t}},-sin{{B}_{t}})dt$$ ${{y}_{1}}=\cos {{B}_{t}},\overset{{}}{\mathop{{}}}\,{{y}_{2}}=\sin {{B}_{t}}$ $$d{{y}_{t}}=(-{{y}_{2}},{{y}_{1}})d{{B}_{t}}+\frac{1}{2}(-{{y}_{1}},-{{y}_{2}})dt\\ $$ or $$\\\left\{ \begin{matrix} d{{y}_{1}}=-{{y}_{2}}d{{B}_{t}}-\frac{1}{2}{{y}_{1}}dt \\ d{{y}_{2}}=+{{y}_{1}}d{{B}_{t}}-\frac{1}{2}{{y}_{2}}dt \\ \end{matrix} \right.\\$$ Matrix form $$\\dY=\frac{-1}{2}Y.dt+K.Y.dB_t$$ K is $$K=\left( \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right)\\$$ .

Khosrotash
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